?2x?1??3x?2?97?9．lim 100x???3x?1?????????3解：原式
8?2x?1??3x?2??2? lim??lim???x??????x??3x?127???3x?1??3?3973x2?ax?610．已知lim存在，则a=
x?11?x解：Qlim?1?x??0?limx?ax?6?0，1?a?6?0,a??7
2x?1x?1???11arcsinx?x11．lim?esin2???
x?0?xx??1111arcsinxx解：Qsin2?1,lim?ex?0?limexsin2?0又Qlim?lim?1，故原式=1.
x?0x?0x?0xx?0xxx12．若limx?0x2ln?1?x2?sinnxx2ln?1?x2?sinnxsinnx?0,则正整数n= ?0且limx?01?cosxxnn?2x2?x2n?40,lim20?n?2,n?4, 故n?3. ?limn
x?0x?0xx2解：Qlimx?013．求lim解：Qlimsin3x?2x
x??sin2x?3xsin3x1sin2x1?????0?sin3x?1,lim?0?lim?0?sin2x?1,lim?0?
x??x??xx??xx??x??x???原式?0?22?? 0?3314．求limx?01?tanx?1?sinx
x?1?cosx?limx?0解：原式
有理化tanx?sinxx(1?cosx)(1?tanx?1?sinx)?limtanx(1?cosx)1?
x?0x(1?cosx)2?limtanx11x1??lim?
x??x22x?0x2x21??15．求lim?sin?cos?
x??xx??解：令
1?t，当x??时，t?0 x1t1t原式?lim?cost?sin2t??lim?1?cost?1?sin2t?t?0t?0???elimcost?1?sin2tt?0t?e2
16．求limlncos2x
x?0lncos3x122x??ln?1?cos2x?1?等价变形cos2x?1等价4lim解：原式limlim2? x?0cos3x?1x?0ln?1?cos3x?1?x?0129??3x?2?注:原式
????????lim4?2sin2xcos3x???? ?x?0cos2x9?3sin3xex?e?x?2x17．求lim
x?0x?sinx 解： 原式
0000ex?e?x0ex?e?xex?e?x?20limlim?2 limx?0x?0cosxx?01?cosxsinx??1x??e?a,x?018．设f?x???且limf?x?存在，求a的值。
x?01?cosx?,x?0?x???1???xe?a解：Qlim???e?a?0?a?a x?0???1x2??x?22?lim2??
x?0?x2xx?0lim?1?cosx?limx?0?x2 211?3lnx?a??19．lim?sin3x??x?0
x?0?sin3x解： 原式
?0?换底法0lim3cosx3xex?0?limln(sin3x)1?3lnx?e?ex?0?3sinxlim3x?ex?0?3xlimx?e
1320．求lim?x?x2ln?1?x??????1???? x??解： 原式
1?tx?0?1??1?t?ln?1?t??0??1ln?1?t??通分1?t limlim??lim?t?0t?0tt?0t22tt2???limt?01?t?111?lim?
2t?t?1?t?0t?12x??21．求lim3x?9x2?12x?1
??解： 原式
有理化lim9x2??9x2?2x?1?3x?9x?2x?12x???lim?2x?13x?9x?2x?12x??
1?21x?lim???.
x??3?33213?9??2xx?2?x2?mx?8122． 已知lim2?，求常数m,n的值。
x?2x??2?n?x?2n5解：（1）∵原极限存在且lim??x??2?n?x?2n???0
2x?2?lim?x2?mx?8??0,4?2m?8?0，2m?12,m?6
x?2x2?6x?8（2）Qlim2x?2x??2?n?x?2n?0????0?lim2x?64?6?21???
x?22x??2?n?4??2?n?2?n5??10?2?n n?12 答m?6,n?12
第三部分：
1．若f?x?为是连续函数，且f?0??1,f?1??0，则limf?xsinx????1???( ) x?A． -1 B．0 C．1 D． 不存在 1??sinx??f?1??0，选B 解： 原式f连续f?limxsin1??f?lim?x??1??x??x?????x??2． 要使f?x??ln?1?kx?在点x?0处连续，应给f?0?补充定义的数值是( ) A． km B．
mxkkm C． lnkm D． e mmmlimkx???x?0xx?lnekm?km 解：Qlimf?x??ln?lim(1?kx)??lnex?0?x?0??f?0??km 选A
3．若limf(x)?A，则下列正确的是 （ ）
x?aA． limf?x??A B． limx?ax?af?x??A C． limf?x???A D． limf(x)?A
x?ax?a解：limx?af?x?u连续limf?x??x?aA 选B
?f?x?,x?0?4．设F?x???x且f?x?在x?0处可导，f??0??0,f?0??0,则x?0是
?f?0?,x?0?F?x?的 ( )
A． 可去间断点 B．跳跃间断点 C．无穷间断点 D． 连续点 解：QlimF?x??limx?0x?0f?x??f?0?x?0x?0?f??0?,
f??0??f?0??F?0??f?0??limF?0?，故x?0是F?x?的第一类可去间断点。选A 1?xsin?5．f?x???x,x?0在x?0处 ( )
??0,x?0A． 极限不存在 B．极限存在但不连续 C ．连续但不可导 D．可导但不连续 解：Qlimf?x??limx?sinx?0x?01?0，且f?0??0 x1xsin?0x?不存在，?f?x?在x?0不可导 ?f?x?在x?0连续，又Qf??0??limx?0x?0选C
?x2?1,x?16．设f?x???在x?1可导，则a,b为 （ ）
?ax?b,x?1A． a??2,b?2 B． a?0,b?2 C． a?2,b?0 D． a?1,b?1
x?1?2,lim解：（1）Qf?x?在x?1连续，?lim?ax?b??a?b ??2x?1x?1??故a?b?2??1?
a?x?1?x2?1ax?b?2?1????2,f??1??lim（2）f??1??limlim?a
x?1?x?1x?1?x?1?x?1x?1?a?2，代入?1?得b?0，选C
7．设f(x)为连续奇函数，则f?0?= 解：（1）Qf?x?为奇函数，?f??x???f?x?
Qlimf??x??lim??f?x??（2）又Qf?x?在x?0连续?f?0???f?0? 故f?0??0 ?，x?0x?0?8．若f?x?为可导的偶函数，则f??0?? 解：（1）Qf?x?为偶函数，?f??x??f?x?
(2)Qf?x?可导，??f???x??f??x? 故?f??0??f??0?2f??0??0 即f??0??0 9．设y?6x?k是曲线y?3x?6x?13的一条切线，则k? 解： （1）Qy??6,y??6x?6,?6x?6?6,x?2 （2）6?2?k?3?4?6?2?13,?12?k?12?12?13,故k?1
2??x??0，则f??0?= 10． 若y?f(x)满足：f(x)?f?0??x???x?，且limx?0x解：f??0??limx?0f?x??f?0?x?0?limx?0x???x?x?1?0?1
4??1?2??
?x?2x?4?11． 设f(x)在x?2连续，且f(2)=4，则limf(x)?x?2解： 原式=f(2)limx?2x?2?411?4lim?4??1 2x?2x?24x?412．f(x)?sinx??x?1?的间断点个数为
x5?x解： 令x5?x?0,x?x?1??x?1?x2?1?0，x?0,x??1,x?1为间断点，
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(完整word版)专升本高数第一章练习题(带答案)
